# call-lim-n-H-n-ln-n-proof-that-is-finite-and-0-1-

Question Number 2783 by 123456 last updated on 27/Nov/15
$$\mathrm{call}\:\gamma:=\underset{{n}\rightarrow+\infty} {\mathrm{lim}H}_{{n}} −\mathrm{ln}\:{n} \\$$$$\mathrm{proof}\:\mathrm{that}\:\gamma\:\mathrm{is}\:\mathrm{finite}\:\mathrm{and}\:\gamma\in\left(\mathrm{0},\mathrm{1}\right) \\$$
Commented by Filup last updated on 27/Nov/15
$$\mathrm{I}\:\mathrm{am}\:\mathrm{curious}\:\mathrm{as}\:\mathrm{to}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{these} \\$$$$\mathrm{kinds}\:\mathrm{of}\:\mathrm{questions}. \\$$
Commented by Filup last updated on 27/Nov/15
$$\mathrm{What}\:\mathrm{does}\::=\:\mathrm{mean}? \\$$$$\mathrm{Same}\:\mathrm{as}\:\equiv\:? \\$$
Commented by 123456 last updated on 27/Nov/15
$$:=\:\mathrm{mean}\:\mathrm{defined} \\$$$$\mathrm{ex}: \\$$$${f}\left({x}\right):={x} \\$$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\$$
Commented by Filup last updated on 27/Nov/15
$${Ah}\:\mathrm{I}\:\mathrm{see}! \\$$
Answered by prakash jain last updated on 27/Nov/15
$$\gamma_{{n}} =\mathrm{H}_{{n}} −\mathrm{ln}\:{n} \\$$$$\mathrm{H}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+..+\frac{\mathrm{1}}{{n}}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} \\$$$${a}_{{i}} =\frac{\mathrm{1}}{{i}} \\$$$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:\mathrm{so}\:\mathrm{that}\:{f}\left({i}\right)={a}_{{i}} \\$$$$\mathrm{Since}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{strictly}\:\mathrm{decreasing}\:+\mathrm{ve}\:\mathrm{function}. \\$$$$\mathrm{From}\:\mathrm{integral}\:\mathrm{test}\:\mathrm{for}\:\mathrm{series} \\$$$$\int_{{N}} ^{{M}+\mathrm{1}} {f}\left({x}\right){dx}\leqslant\underset{{n}={N}} {\overset{{M}} {\sum}}\:{f}\left({n}\right)\leqslant{f}\left({N}\right)+\int_{{N}} ^{{M}} {f}\left({x}\right){dx}\:\:…\left({A}\right) \\$$$$\mathrm{H}_{{n}−\mathrm{1}} \:\geqslant\int_{\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{x}}{dx}=\mathrm{ln}\:{n} \\$$$${S}\mathrm{o}\:\gamma_{{n}} =\mathrm{H}_{{n}} −\mathrm{ln}\:{n}=\frac{\mathrm{1}}{{n}}+\mathrm{H}_{{n}−\mathrm{1}} −\mathrm{ln}\:{n}>\mathrm{0}\:\:\:\:\:\:\:…\left(\mathrm{1}\right) \\$$$$\gamma_{{n}+\mathrm{1}} =\gamma_{{n}} +\frac{\mathrm{1}}{{n}+\mathrm{1}}−\mathrm{ln}\:\left({n}+\mathrm{1}\right)+\mathrm{ln}\:{n} \\$$$$\frac{\mathrm{1}}{{n}+\mathrm{1}}\leqslant\mathrm{ln}\frac{{n}+\mathrm{1}}{{n}}\:\left({comparing}\:{area}\:{of}\:{rectangles}\right) \\$$$$\Rightarrow\gamma_{{n}+\mathrm{1}} =\gamma_{{n}} −\left[\mathrm{ln}\:\frac{{n}+\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right]<\gamma_{{n}} \\$$$$\gamma_{{n}} >\mathrm{0}\:{and}\:\gamma_{{n}+\mathrm{1}} <\gamma_{{n}} \\$$$$\mathrm{So}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\gamma_{{n}} \:{exists}\:\mathrm{and}\:>\mathrm{0}. \\$$$$\gamma_{\mathrm{1}} =\mathrm{1}−\mathrm{ln}\:\mathrm{1}=\mathrm{1} \\$$$$\because\gamma_{{n}+\mathrm{1}} <\gamma_{{n}} \\$$$$\mathrm{0}<\gamma<\mathrm{1} \\$$
Commented by RasheedAhmad last updated on 29/Nov/15
$${What}\:{is}\:\mathrm{H}_{{n}} ? \\$$
Commented by 123456 last updated on 29/Nov/15
$$\mathrm{harmonic}\:\mathrm{numbers} \\$$
Commented by Rasheed Soomro last updated on 29/Nov/15
$$\mathscr{THANKS}! \\$$